At what point would an elementary generalization of Bertrand's Postulate be interesting?

Current results are able to yield such results. Depending on how generous one is regarding what $X$ is. If it is just the optimal value can be calculated exactly this will work for many more $k$ and if one is happy with an explicit bound for all $k$.

For example Dusart showed that $$ \frac{x}{\log x - 1} \le \pi(x) \le \frac{x}{\log x - 1.1} $$ for $x\ge 60184$. Now for some $k$, write $y=kx$. Then, if the upper bound for $kx=y$ is smaller than the lower bound for $(k+1)x = (1+1/k)y$, that is $$ \frac{y}{\log y - 1.1} \lt \frac{y(1+ 1/k)}{\log( y (1+1/k) )- 1} $$ one has a prime between $kx$ and $(k+1)x$, since then $\pi(kx) \lt \pi((k+1)x)$.

One can check that this inequality holds for (up to potential error in my calculation) $$ y \ge 10 e^{0.1 k}. $$

So, for $x \ge \max \lbrace 10 e^{0.1 k}/k , 60184/k \rbrace $ one always has a prime between $kx$ and $(k+1)x$.

While this grows exponential in $k$, the growth is such that it is well feasible to check 'everything' up to the bound to get an optimal $X$ for not too large $k$. And, one always has an explict value.

This proof is of course not elementary (the non-elementariness being hidden in Dusart's result) and is an application of the PNT in some sense. But what this is meant to show is that for a result around this to be interesting it seems necessary either to be better (and one could still optimize this here) than this or the proof would have to be interesting (or both). [What an interesting proof is is of course a bit subjective.]

I think that it would be interesting if it has an effective (and not too huge) value of $X$.

Once again, being new here I tried to comment your answer, but obviously don't have 50 points yet.

Here is the original text of my comment with my mistake on value of x

"As a check on your equations I set k=100 x=2478 and should therefore expect one prime between 2478 and 2478 + 2478/100 = 2502.78.(for k=100, x must be greater than 2202.65 as detailed by your equations - this condition is met for x=2478).But there are no primes between 2478 and 2502.78. (Though 2503 is prime) Is the domain of x for your answer the positive integers? Then everything checks out okay. Or as you warned 'up to error'?"

Assigning x=24.78 is the correct value.

@quid: Note the prime is between kx and (k+1)x, so with my corrected value of x there should be a prime between 2478 < (some prime) < 2502.78, but the next prime is 2503.

@stefan kohl: understood and my calculation and answer/comment also gave x > 2202.65. My mistake was not putting the correct value for x